CODEWITHSUNDEEP
listhaskellfilter
Cem Atalay
Cem Atalay

Reputation: 71

A Haskell function that takes two strings and filters the second according to the first

The objective is: Using foldr, define a function remove which takes two strings as its arguments and removes every letter from the second list that occurs in the first list. For example, remove "first" "second" = "econd".

If this function took a single character and a string, I would do:

remove a xs = foldr (\x acc -> if x /= a then x : acc else acc) [] xs

But I can't figure out how I am supposed to do this with two strings. Thank you!

Upvotes: 2

Views: 721

Answers (2)

dopamane
dopamane

Reputation: 1315

remove :: String -> String -> String                       
remove xs ys = foldr (condCons) "" ys
  where
    condCons l rs | l `notElem` xs = l : rs
                  | otherwise = rs

It is also allowed to discard the 'ys' paramenter:

remove :: String -> String -> String                       
remove xs = foldr (condCons) ""
  where
    condCons l rs | l `notElem` xs = l : rs
                  | otherwise = rs

Basically, condCons takes a character L and a string Rs. If L is not an element of xs, then it cons to Rs, otherwise leave Rs unchanged. foldr takes the condCons, the intial string "", and the second argument ys. L takes on each character of the string ys from right to left, building a new string from "" using the condCons binary operator.

Upvotes: 1

Cem Atalay
Cem Atalay

Reputation: 71

remove xs ys = foldr (\x acc -> if elem x xs then acc else x : acc) [] ys

yep.

Upvotes: 5

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